Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 917    Accepted Submission(s): 434

Problem Description
As is known to all, in many cases, a word has two meanings. Such as “hehe”, which not only means “hehe”, but also means “excuse me”. 
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.
 
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.

Limits
T <= 30
|A| <= 100000
|B| <= |A|

 
Output
For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 1000000007.
 
Sample Input
4
hehehe
hehe
woquxizaolehehe
woquxizaole
hehehehe
hehe
owoadiuhzgneninougur
iehiehieh
 
Sample Output
Case #1: 3 Case #2: 2 Case #3: 5 Case #4: 1

Hint

In the first case, “ hehehe” can have 3 meaings: “*he”, “he*”, “hehehe”. In the third case, “hehehehe” can have 5 meaings: “*hehe”, “he*he”, “hehe*”, “**”, “hehehehe”.

题意:给你一个主串一个子串,然后主串中匹配到子串就可以把匹配部分改为*(也可以不改),问主串有多少钟不同的样子;

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e5+10; int la,lb;
ll dp[N];
char a[N],b[N];
ull ha[N],hb;
ull seed=13331; void init(){
hb=0;
for(int i=0;i<lb;i++)
hb=hb*seed+b[i];
ull base=1;
for(int i=1;i<=lb-1;i++) base*=seed;
ha[0]=a[0];
for(int i=1;i<=lb-1;i++)
ha[i]=ha[i-1]*seed+a[i];
for(int i=lb;i<la;i++)
ha[i]=(ha[i-1]-a[i-1-(lb-1)]*base)*seed+a[i];
} int main()
{
int cas,kk=0;
scanf("%d",&cas);
while(cas--){
scanf("%s",a);
scanf("%s",b);
la=strlen(a);lb=strlen(b);
if(la<lb) {printf("Case #%d: 1\n",++kk);CT;}
init();
for(int i=0;i<=lb-1;i++) dp[i]=1;
if(ha[lb-1]==hb) dp[lb-1]=2;
for(int i=lb;i<la;i++){
dp[i]=dp[i-1]%mod;
if(ha[i]==hb) dp[i]=(dp[i]+dp[i-lb])%mod;
}
printf("Case #%d: %lld\n",++kk,dp[la-1]%mod);
}
return 0;
}

  分析:错误点:

1.BKDRhash不太熟练,只会原来的最后输出&的形式,导致最后计算复杂;

改进:BKDRhash的形式:pre*seed+a[i],seed为13331之类的大素数,pre为i以前的哈希值;

a[i]就是字符

2,没有想到dp,当前下标i的话,dp[i]的数值,如果当前i并未匹配到子串,dp[i]=dp[i-1];

如果匹配到子串,dp[i]=dp[i-1]+dp[i-lb];

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